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Beginning and Intermediate Algebra

Chapter 8: Radicals

An open source (CC-BY) textbook

by Tyler Wallace

1

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2

Chapter 8: Radicals

8.1

Radicals - Square Roots

Square roots are the most common type of radical used. A square root “un-squares” a number. For example, because 52 = 25 we say the square root of 25 is 5.The square root of 25 is written as 25

√. The following example gives several

square roots:

Example 1.

1√

= 1 121√

= 11

4√

= 2 625√

= 25

9√

= 3 − 81√

=Undefined

The final example, − 81√

is currently undefined as negatives have no square root.This is because if we square a positive or a negative, the answer will be positive.Thus we can only take square roots of positive numbers. In another lesson we willdefine a method we can use to work with and evaluate negative square roots, butfor now we will simply say they are undefined.

Not all numbers have a nice even square root. For example, if we found 8√

onour calculator, the answer would be 2.828427124746190097603377448419... andeven this number is a rounded approximation of the square root. To be as accu-rate as possible, we will never use the calculator to find decimal approximations ofsquare roots. Instead we will express roots in simplest radical form. We will dothis using a property known as the product rule of radicals

ProductRule of SquareRoots: a · b√

= a√

· b√

We can use the product rule to simplify an expression such as 36 · 5√

by spliting

3

it into two roots, 36√

· 5√

, and simplifying the first root, 6 5√

. The trick in this

process is being able to translate a problem like 180√

into 36 · 5√

. There are sev-eral ways this can be done. The most common and, with a bit of practice, thefastest method, is to find perfect squares that divide evenly into the radicand, ornumber under the radical. This is shown in the next example.

Example 2.

75√

75 is divisible by 25, a perfect square

25 · 3√

Split into factors

25√

· 3√

Product rule, take the square root of 25

5 3√

Our Solution

If there is a coefficient in front of the radical to begin with, the problem mearlybecomes a big multiplication problem.

Example 3.

5 63√


5 9 · 7√

Split into factors

5 9√

· 7√


5 · 3 7√

Multiply coefficients

15 7√

Our Solution

As we simplify radicals using this method it is important to be sure our finalanswer can be simplifyied no more.

Example 4.

72√


9 · 8√

Split into factors

9√

· 8√


3 8√

But 8 is also divisible by a perfect square, 4

3 4 · 2√

Split into factors

3 4√

· 2√


4

3 · 2 2√

Multiply

6 2√

Our Solution.

The previous example could have been done in fewer steps if we had noticed that72= 36 · 2, but often the time it takes to discover the larger perfect square is morethan it would take to simplify in several steps.

Variables often are part of the radicand as well. When taking the square roots of

variables, we can divide the exponent by 2. For example, x8√

= x4, because wedivide the exponent of 8 by 2. This follows from the power of a power rule ofexpoennts, (x4)2 = x8. When squaring, we multiply the exponent by two, so whentaking a square root we divide the exponent by 2. This is shown in the followingexample.

Example 5.

− 5 18x4y6z10√


− 5 9 · 2x4y6z10√

Split into factors

− 5 9√

· 2√

· x4√

· y6√

· z10√

Product rule, simplify roots, divide exponents by 2

− 5 · 3x2y3z5 2√


− 15x2y3z5 2√

Our Solution

We can’t always evenly divide the exponent on a variable by 2. Sometimes wehave a remainder. If there is a remainder, this means the remainder is left insidethe radical, and the whole number part is how many are outside the radical. Thisis shown in the following example.

Example 6.

20x5y9z6√


4 · 5x5y9z6√

Split into factors

4√

· 5√

· x5√

· y9√

· z6√

Simplfy,divide exponents by 2, remainder is left inside

2x2y4z3 5xy√

Our Solution

5

Practice - Square Roots

Simplify.

1) 245√

3) 36√

5) 12√

7) 3 12√

9) 6 128√

11) − 8 392√

13) 192n√

15) 196v2√

17) 252x2√

19) − 100k4√

21) − 7 64x4√

23) − 5 36m√

25) 45x2y2√

27) 16x3y3√

29) 320x4y4√

31) 3 320x4y4√

33) 5 245x2y3√

35) − 2 180u3v√

37) − 8 180x4y2z4√

39) 2 80hj4k√

41) − 4 54mnp2√

2) 125√

4) 196√

6) 72√

8) 5 32√

10) 7 128√

12) − 7 63√

14) 343b√

16) 100n3√

18) 200a3√

20) − 4 175p4√

22) − 2 128n√

24) 8 112p2√

26) 72a3b4√

28) 512a4b2√

30) 512m4n3√

32) 8 98mn√

34) 2 72x2y2√

36) − 5 72x3y4√

38) 6 50a4bc2√

40) − 32xy2z3√

42) − 8 32m2p4q√

6

8.2

Radicals - Higher Roots

While square roots are the most common type of radical we work with, we cantake higher roots of numbers as well: cube roots, fourth roots, fifth roots, etc. Fol-lowing is a definition of radicals.

am√

= b if bm= a

The small letter m inside the radical is called the index. It tells us which root weare taking, or which power we are “un-doing”. For square roots the index is 2. Asthis is the most common root, the two is not usually written. The followingexample includes several higher roots.

Example 7.

1253√

=5 − 643√

=− 4

814√

= 3 − 1287√

=− 2

325√

= 2 − 164√

=undefined

We must be careful of a few things as we work with higher roots. First its impor-

tant not to forget to check the index on the root. 81√

= 9 but 814√

= 3. This isbecause 92 = 81 and 34 = 81. Another thing to watch out for is negatives underroots. We can take an odd root of a negative number, because a negative numberraised to an odd power is still negative. However, we cannot take an even root ofa negative number, this we will say is undefined. In a later section we will discusshow to work with roots of negative, but for now we will simply say they are unde-fined.

We can simplify higher roots in much the same way we simplified square roots,using the product property of radicals.

Product Property ofRadicals: abm√

= am√

· bm√

Often we are not as familiar with higher powers as we are with squares. It isimportant to remember what index we are working with as we try and work ourway to the solution.

Example 8.

543√

Weareworkingwith a cubed root,want third powers

23 = 8 Test 2, 23 =8, 54 is not divisible by 8.

33 = 27 Test 3, 33 = 27, 54 is divisible by 57!

27 · 23√

Write as factors

273√

· 23√

Product rule, take cubed root of 27

3 23√

Our Solution

7

Just as with square roots, if we have a coefficient, we multiply the new coefficientstogether.

Example 9.

3 484√

Weareworkingwith a fourth root,want fourth powers

24 = 16 Test 2, 24 = 16, 48 is divisible by 16!

3 16 · 34√

Write as factors

3 164√

· 34√

Product rule, take fourth root of 16

3 · 2 34√


6 34√

Our Solution

We can also take higher roots of variables. As we do, we will divide the exponenton the variable by the index. Any whole answer is how many of that varible willcome out of the square root. Any remainder is how many are left behind insidethe square root. This is shown in the following examples.

Example 10.

x25y17z35√

Divide each exponent by 5,whole number outside, remainder inside

x5y3 y2z35√

Our Solution

In the previous example, for the x, we divided25

5= 5R 0, so x5 came out, no x’s

remain inside. For the y, we divided17

5= 3R2, so y3 came out, and y2 remains

inside. For the z, when we divided3

5= 0R3, all three or z3 remained inside. The

following example includes integers in our problem.

Example 11.

2 40a4b83√

Looking for cubes that divide into 40.The number 8works!

2 8 · 5a4b83√

Take cube root of 8, dividing exponents on variables by 3

2 · 2ab2 5ab23√

Remainders are left in radical.Multiply coefficients

4ab2 5ab23√

Our Solution

8

Practice - Higher Roots

Simplify.

1) 6253√

3) 7503√

5) 8753√

7) − 4 964√

9) 6 1124√

11) − 1124√

13) 648a24√

15) 224n35√

17) 224p55√

19) − 3 896r7√

21) − 2 − 48v73√

23) − 7 320n63√

25) − 135x5y33√

27) − 32x4y43√

29) 256x4y63√

31) 7 − 81x3y73√

33) 2 375u2v83√

35) − 3 192ab23√

37) 6 − 54m8n3p73√

39) 6 648x5y7z24√

41) 7 128h6j8k84√

2) 3753√

4) 2503√

6) 243√

8) − 8 484√

10) 3 484√

12) 5 2434√

14) 64n34√

16) − 96x35√

18) 256x66√

20) − 8 384b87√

22) 4 250a63√

24) − 512n63√

26) 64u5v33√

28) 1000a4b53√

30) 189x3y63√

32) − 4 56x2y83√

34) 8 − 750xy3√

36) 3 135xy33√

38) − 6 80m4p7q44√

40) − 6 405a5b8c4√

42) − 6 324x7yz74√

9

8.3

Radicals - Add/Subtract Radicals

Adding and subtracting radicals is very similar to adding and subtracting withvariables. Consider the following example.

Example 12.

5x + 3x− 2x Combine like terms

6x Our Solution

5 11√

+ 3 11√

− 2 11√

Combine like terms

6 11√

Our Solution

Notice that when we combined the terms with 11√

it was just like combining

10

terms with x. When adding and subtracting with radicals we can combine likeradicals just like like terms. We add and subtract the coefficients in front of theradical, and the radical stays the same. This is shown in the following example.

Example 13.

7 65√

+ 4 35√

− 9 35√

+ 65√

Combine like radicals 7 65√

+ 65√

and 4 35√

− 8 35√

8 65√

− 5 35√

Our Solution

We cannot simplify this expression any more as the radicals do not match. Oftenproblems we solve have no like radicals, however, if we simplify the radicals firstwe may find we do in fact have like radicals.

Example 14.

5 45√

+6 18√

− 2 98√

+ 20√

Simplify radicals, findperfect square factors

5 9 · 5√

+6 9 · 2√

− 2 49 · 2√

+ 4 · 5√

Take rootswhere possible

5 · 3 5√

+ 6 · 3 2√

− 2 · 7 2√

+ 2 5√


15 5√

+ 18 2√

− 14 2√

+ 2 5√

Combine like terms

17 5√

+ 4 2√

Our Solution

This exact process can be used to add and subtract radicals with higher indicies

Example 15.

4 543√

− 9 163√

+5 93√

Simplify each radical, finding perfect cube factors

4 27 · 23√

− 9 8 · 23√

+5 93√

Take rootswhere possible

4 · 3 23√

− 9 · 2 23√

+5 93√


12 23√

− 18 23√

+5 93√

Combien like terms 12 23√

− 18 23√

− 6 23√

+5 93√

Our Solution

11

Practice - Add and Subtract Radicals

Simiplify

1) 2 5√

+ 2 5√

+ 2 5√

3) − 3 2√

+3 5√

+3 5√

5) − 2 6√

− 2 6√

− 6√

7) 3 6√

+ 3 5√

+ 2 5√

9) 2 2√

− 3 18√

− 2√

11) − 3 6√

− 12√

+ 3 3√

13) 3 2√

+ 2 8√

− 3 18√

15) 3 18√

− 2√

− 3 2√

17) − 3 6√

− 3 6√

− 3√

+ 3 6√

19) − 2 18√

− 3 8√

− 20√

+2 20√

21) − 2 24√

− 2 6√

+ 2 6√

+ 2 20√

23) 3 24√

− 3 27√

+2 6√

+2 8√

25) − 2 163√

+2 163√

− 2 23√

27) 2 2434√

− 2 2434√

− 34√

29) 3 24√

− 2 24√

− 2434√

31) − 3244√

+3 3244√

− 3 44√

33) 2 24√

+ 2 34√

+ 3 644√

− 34√

35) − 3 65√

− 645√

+ 2 1925√

− 2 1605√

37) 2 1605√

− 2 1925√

− 1605√

− − 1605√

39) − 2566√

− 2 46√

− 3 3206√

− 2 1286√

2) − 3 6√

− 3 3√

− 2 3√

4) − 2 6√

− 3√

− 3 6√

6) − 3 3√

+2 3√

− 2 3√

8) − 5√

+2 3√

− 2 3√

10) − 54√

− 3 6√

+ 3 27√

12) − 5√

− 5√

− 2 54√

14) 2 20√

+ 2 20√

− 3√

16) − 3 27√

+2 3√

− 12√

18) − 2 2√

− 2√

+ 3 8√

+ 3 6√

20) − 3 18√

− 8√

+ 2 8√

+ 2 8√

22) − 3 8√

− 5√

− 3 6√

+2 18√

24) 2 6√

− 54√

− 3 27√

− 3√

26) 3 1353√

− 813√

− 1353√

28) − 3 44√

+3 3244√

+2 644√

30) 2 64√

+ 2 44√

+ 3 64√

32) − 2 2434√

− 964√

+ 2 964√

34) 2 484√

− 3 4054√

− 3 484√

− 1624√

36) 2 2565√

− 2 2567√

− 3 27√

− 6407√

38) − 2 2567√

− 2 2567√

− 3 27√

− 6407√

40) − 3 37√

− 3 7687√

+ 2 3847√

+ 3 57√

12

8.4

Radicals - Multiply and Divide Radicals

Multiplying radicals is very simple if the index on all the radicals match. Theprodcut rule of radicals which we have alread been using can be generalized as fol-lows:

ProductRule ofRadicals: a bm√

· c dm√

= ac bdm√

Another way of stating this rule is we are allowed to multiply the factors outsidethe radical and we are allowed to multiply the factors inside the radicals, as longas the index matches. This is shown in the following example.

Example 16.

− 5 14√

· 4 6√

Multiply outside and inside the radical

− 20 84√

Simplify the radical,divisible by 4

− 20 4 · 21√

Take the square rootwhere possible

− 20 · 2 21√


− 40 21√

Our Solution

The same process works with higher roots

Example 17.

2 183√

· 6 153√

Multiply outside and inside the radical

12 2703√

Simplify the radical, divisible by 27

12 27 · 103√

Take cube rootwhere possible

12 · 3 103√


36 103√

Our Solution

When multiplying with radicals we can still use the distributive property or FOILjust as we could with variables.

Example 18.

7 6√

(3 10√

− 5 15√

) Distribute, following rules formultiplying radicals

21 60√

− 35 90√

Simplify each radical, finding perfect square factors

21 4 · 15√

− 35 9 · 10√

Take square rootwhere possible

21 · 2 15√

− 35 · 3 10√


42 15√

− 105 10√

Our Solution

13

Example 19.

( 5√

− 2 3√

)(4 10√

+ 6 6√

) FOIL, following rules formultiplying radicals

4 50√

+ 6 30√

− 8 30√

− 12 18√

Simplify radicals, find perfect square factors

4 25 · 2√

+6 30√

− 8 30√

− 12 9 · 2√

Take square rootwhere possible

4 · 4 2√

+6 30√

− 8 30√

− 12 · 3 2√


16 2√

+ 6 30√

− 8 30√

− 36 2√

Combine like terms

− 20 2√

− 2 30√

Our Solution

Example 20.

(2 5√

− 3 6√

)(7 2√

− 8 7√

) FOIL, following rules formultiplying radicals

14 10√

− 16 35√

− 21 12√

− 24 42√

Simplify radicals,find perfect square factors

14 10√

− 16 35√

− 21 4 · 3√

− 24 42√

Take square root where possible

14 10√

− 16 35√

− 21 · 2 3√

− 24 42√

Multiply coefficient

14 10√

− 16 35√

− 42 3√

− 24 42√

Our Solution

As we are multiplying we always look at our final solution to check if all the radi-cals are simplified and all like radicals or like terms have been combined.

Division with radicals is very similar to multiplication, if we think about divisionas reducing fractions, we can reduce the coefficients outside the radicals andreduce the values inside the radicals to get our final solution.

QuotientRule ofRadicals:a b

m√

c dm√ =

a

c

b

d

m

√

Example 21.

15 1083√

20 23√ Reduce

15

20and

1083√

2√ by dividing by 5 and 2 respectively

3 543√

4Simplify radical, 54 is divisible by 27

3 27 · 23√

4Take the cube root of 27

3 · 3 23√

4Multiply coefficients

9 23√

4Our Solution

There is one catch to dividing with radicals, it is considered bad practice to havea radical in the denominator of our final answer. If there is a radical in thedenominator we will rationalize it, or clear out any radicals in the denominator.

14

We do this by multiplying the numerator and denominator by the same thing.The problems we will consider here will all have a monomial in the denominator.The way we clear a monomial radical in the denominator is to focus on the index.The index tells us how many of each factor we will need to clear the radical. Forexample, if the index is 4, we will need 4 of each factor to clear the radical. Thisis shown in the following examples.

Example 22.

6√

5√ Index is 2,weneed two fives in denominator, need 1more

6√

5√

(

5√

5√

)

Multiply numerator and denominator by 5√

30√

5Our Solution

Example 23.

3 114√

24√ Index is 4,we need four twos in denominator, need 3more

3 114√

24√

(

234√

234√

)


3 884√

2Our Solution

Example 24.

4 23√

7 253√ The 25 can bewritten as 52.Thiswill help us keep the numbers small

4 23√

7 523√ Index is 3,we need three fives in denominator, need 1more

4 23√

7 523√

(

53√

53√

)


4 103√

7 · 5Multiply out denominator

4 103√

35Our Solution

15

The previous example could have been solved by multiplying numerator and

denominator by 2523√

. However, this would have made the numbers very largeand we would have needed to reduce our soultion at the end. This is why re-writing the radical as 523

√and multiplying by just 53

√was the better way to sim-

plify.

We will also always want to reduce our fractions (inside and out of the radical)before we rationalize.

Example 25.

6 14√

12 22√ Reduce coefficients and inside radical

7√

2 11√ Index is 2, need two elevens, need 1more

7√

2 11√

(

11√

11√

)

Multiply numerator anddenominator by 11√

77√

2 · 11Multiply denominator

77√

22Our Solution

The same process can be used to rationalize fractions with variables.

Example 26.

18 6x3y4z4√

8 10xy6z34√ Reduce coefficients and inside radical

9 3x24√

4 5y2z34√

Index is 4.Weneed four of everything to rationalize,threemore fives, twomore y ′s and onemore z.

9 3x24√

4 5y2z34√

(

53y2z4√

53y2z4√

)

Multiply numerator and denominator by 53y2z4√

9 375x2y2z4√

4 · 5yzMultiply denominator

9 375x2y2z4√

20yzOur Solution

16

Practice - Multiply and Divide Radicals

State if the given functions are inverses.

1) 3 5√

·− 4 16√

3) 12m√

· 15m√

5) 4x33√

· 2x43√

7) 6√

( 2√

+2)

9) − 5 15√

(3 3√

+2)

11) 5 10√

(5n + 2√

)

13) (2+ 2 2√

)(− 3 + 2√

)

15) ( 5√

− 5)(2 5√

− 1)

17) ( 2a√

+2 3a√

)(3 2a√

+ 5a√

)

19) (− 5− 4 3√

)(− 3− 4 3√

)

21)12

√

5 100√

23)5

√

4 125√

25)10

√

6√

27)2 4√

3 3√

29)5x2

4 3x3y3√

31)2p2

√

3p√

33)3 103√

5 273√

35)53

√

4 43√

37)5 5r44√

8r24√

2) − 5 10√

· 15√

4) 5r3√

·− 5 10r2√

6) 3 4a43√

· 10a33√

8) 10√

( 5√

+ 2√

)

10) 5 15√

(3 3√

+2)

12) 15√

( 5√

− 3 3v√

)

14) (− 2+ 3√

)(− 5+ 2 3√

)

16) (2 3√

+ 5√

)(5 3√

+2 4√

)

18) (− 2 2p√

+5 5√

)( 5p√

+ 5p√

)

20) (5 2√

− 1)(− 2m√

+ 5)

22)15

√

2 4√

24)12

√

3√

26)2

√

3 5√

28)4 3√

15√

30)4

5 3xy4√

32)8n2

√

10n√

34)15

3√

643√

36)24

√

2 644√

38)4

65m4n24√

17

8.5

Radicals - Rationalize Denominators

It is considered bad practice to have a radical in the denominator of a fraction.When this happens we multiply the numerator and denominator by the samething in order to clear the radical. In the lesson on dividing radicals we talkedabout how this was done with monomials. Here we will look at how this is donewith binomials.

If the binomial is in the numerator the process to rationalize the denominator isessentially the same as with monomials. The only difference is we will have to dis-tribute in the numerator.

Example 27.

3√

− 9

2 6√ Want to clear 6

√in denominator,multiply by 6

√

( 3√

− 9)

2 6√

(

6√

6√

)

Wewill distribute the 6√

through the numerator

18

18√

− 9 6√

2 · 6Simplify radicals in numerator,multiply out denominator

9 · 2√

− 9 6√

12Take square rootwhere possible

3 2√

− 9 6√

12Reduce by dividing each termby 3

2√

− 3 6√

4Our Solution

It is important to remember that when reducing the fraction we cannot reducewith just the 3 and 12 or just the 9 and 12. When we have addition or subtrac-tion in the numerator or denominator we must divide all terms by the samenumber.

The problem can often be made easier if we first simplify any radicals in theproblem.

2 20x5√

− 12x2√

18x√ Simplify radicals by finding perfect squares

2 4 · 5x3√

− 4 · 3x2√

9 · 2x√ Simplify roots,divide exponents by 2.

2 · 2x2 5x√

− 2x 3√

3 2x√ Multiply coefficients

4x2 5x√

− 2x 3√

3 2x√ Multiplying numerator and denominator by 2x

√

(4x2 5x√

− 2x 3√

)

3 2x√

(

2x√

2x√

)

Distribute through numerator

4x2 10x2√

− 2x 6x√

3 · 2xSimplify roots in numerator,multiply coefficients in denominator

4x3 10√

− 2x 6x√

6xReduce,dividing each termby 2x

19

2x2 10√

− 6x√

3xOur Solution

As we are rationalizing it will always be important to constantly check ourproblem to see if it can be simplified more. We ask ourselves, can the fraction bereduced? can the radicals be simplified? These steps may happen several times onour way to the solution.

If the binomial occurs in the denominator we will have to use a different strategyto clear the radical. Consider

2

3√

− 5, if we were to multiply the denominator by

3√

we would have to distribute it and we would end up with 3 − 5 3√

. Wehaven’t cleared the radical, only moved it to another part of the denominator. Soour current method will not work. Instead we will use what is called a conjuage.A conjugate is made up of the same terms, with the opposite sign in the middle.

So for our example with 3√

− 5 in the denominator, the conjugate would be 3√

+5. The advantage of a conjugate is when we multiply them together we have

( 3√

− 5)( 3√

+ 5), which is a sum and a difference. We know when we multiplythese we get a difference of squares. Squaring 3

√and 5, with subtraction in the

middle gives the product 3 − 25 = − 22. Our answer when multiplying conjugateswill no longer have a square root. This is exactly what we want.

Example 28.

2

3√

− 5Multiply numerator anddenominator by conjugate

2

3√

− 5

(

3√

+5

3√

+5

)

Distribute numerator,difference of squares in denominator

2 3√

+ 10

3− 25Simplify denoinator

2 3√

+ 10

− 22Reduce by dividing all terms by− 2

− 3√

− 5

11Our Solution

In the previous example, we could have reduced by dividng by 2, giving the solu-

tion3

√+ 5

− 11, both answers are correct.

Example 29.

15√

5√

+ 3√ Multilpy by conjugate, 5

√− 3√

20

15√

5√

+ 3√

(

5√

− 3√

5√

− 3√

)

Distribute numerator,denominator is difference of squares

75√

− 45√

5− 3Simplify radicals in numerator, subtract in denominator

25 · 3√

− 9 · 5√

2Take square rootswhere possible

5 3√

− 3 5√

2Our Solution

Example 30.

2 3x√

4− 5x3√ Multiply by conjugate, 4+ 5x3

√

2 3x√

4− 5x3√

(

4+ 5x3√

4+ 5x3√

)

Distribute numerator, denominator is difference of squares

8 3x√

+ 2 15x4√

16− 5x3Simplify radicals where possible

8 3x√

+ 2x2 15√

16− 5x3Our Solution

The same process can be used when there is a binomial in the numerator anddenominator. We just need to remember to FOIL out the numerator.

Example 31.

3− 5√

2− 3√ Multiply by conjugate, 2+ 3

√

3− 5√

2− 3√

(

2+ 3√

2+ 3√

)

FOIL in numerator,denominator is difference of squares

6+ 3 3√

− 2 5√

− 15√

4− 3Simplify denominator

6+ 3 3√

− 2 5√

− 15√

1Divide each termby 1

21

6+ 3 3√

− 2 5√

− 15√

Our Solution

Example 32.

2 5√

− 3 7√

5 6√

+ 4 2√ Multiply by the conjugate, 5 6

√− 4 2

√

2 5√

− 3 7√

5 6√

+ 4 2√

(

5 6√

− 4 2√

5 6√

− 4 2√

)

FOIL numerator,denominator is difference of squares

10 30√

− 8 10√

− 15 42√

+ 12 14√

25 · 6− 16 · 2Multiply in denominator

10 30√

− 8 10√

− 15 42√

+ 12 14√

150− 32Subtract in denominator

10 30√

− 8 10√

− 15 42√

+ 12 14√

118Our Solution

The same process is used when we have variables

Example 33.

3x 2x√

+ 4x3√

5x− 3x√ Multiply by the conjugate, 5x + 3x

√

3x 2x√

+ 4x3√

5x− 3x√

(

5x + 3x√

5x + 3x√

)

FOIL in numerator,denominator is difference of squares

15x2 2x√

+3x 6x2√

+5x 4x3√

+ 12x4√

25x2− 3xSimplify radicals

15x2 2x√

+ 3x2 6√

+ 10x2 x√

+ 2x2 3√

25x2− 3xDivide each termby x

15x 2x√

+3x 6√

+ 10x x√

+2x 3√

25x− 3Our Solution

22

Practice - Rationalize Denominators

Simplify.

1)4+ 2 3

√

9√

3)4+ 2 3

√

5 4√

5)2− 5 5

√

4 13√

7)2

√− 3 3

√

3√

9)2p + 3 5p4

√

5 20p2√

11)3m2

√− 4 2m4

√

5 12m4√

13)5

3 5√

+ 2√

15)2

5+ 2√

17)3

4− 3 3√

19)4

3+ 5√

21) −4

4− 4 2√

23)5

n4√

− 5

25)4p

3− 5 p4√

27)4

5+ 5x2√

29)5

2+ 5r3√

31)5

− 5v − 3 v√

33)4 2√

+ 3

3 2√

+ 3√

35)2− 5

√

− 3+ 5√

37)5 2√

+ 3√

5+ 5 2√

39)3

√+ 2

√

2 3√

− 2√

41)3

√− 2

√

4+ 5√

43)4+ 2 2x2

√

5+ 2 5x3√

45)2 3m2√

− 2m4√

5− 3m2√

47)2b− 5 2b

√

− 1+ 2b4√

49)2− 2x

√

4x − 5 3x3√

51)− 4p− p

√

− p− p3√

2)− 4+ 3

√

4 9√

4)2 3√

− 2

2 16√

6)5

√+ 4

4 17√

8)5

√− 2

√

3 6√

10)5x − 3x4

√

2 19x2√

12)3 2r4√

− 3r4√

14r3√

14)5

3√

+ 4 5√

16)5

2 3√

− 2√

18)4

2√

− 2

23

20)2

2 5√

+ 2 3√

22)4

4 3√

− 5√

24)3n3

√

− 5− 2n4√

26)5x2

5− 3 5x√

28)4b3

− 4b4− 2 b4√

30)5

3a4√

+ 4a

32)4

2 2n2√

+ n3√

34)− 5− 4 5

√

5− 5 3√

36)− 1+ 5

√

2 5√

+ 5 2√

38)4 2√

+ 2 3√

5√

− 4

40)3

√− 5

√

4− 2√

42)2√

2− 3√

5√

− 3 3√

44)3 p4√

+ 5p3√

4− 2p4√

46)4v − 2 2v2

√

− 4− 5 v√

48)− 4n − 3n2

√

− 3− n4√

50)3a +2 2a2

√

3− 5 3a2√

52)4− 5x

√

− 2x− 3x2√

24

8.6

Radicals - Rational Exponents

When we simplify radicals with exponents, we divide the exponent by the index.Another way to write division with with a fraction bar. This idea is how we willdefine rational exponents.

Definition ofRational Exponents: an

m = ( am√

)n

The denominator of a rational exponent becomes the index on our radical, like-wise the index on the radical becomes the denominator of the exponent. We canuse this property to change any radical expression into an exponential expression.

Example 34.

( x5√

)3 = x5

5 ( 3x6√

)5 = (3x)5

6

1

( a7√

)3= a

− 3

71

( xy3√

)2=(xy)

− 2

3

Index is denominatorNegative exponents from reciprocals

We can also change any rational expoennt into a radical expression by using thedenominator as the index.

Example 35.

a5

3 = ( a3√

)5 (2mn)2

7 =( 2mn7√

)2

x− 4

5 =1

( x5√

)4(xy)

− 2

9 =1

( xy9√

)2

Index is denominatorNegative exponentmeans reciprocals

The ability to change between exponential expressions and radical expressionsallows us to evaluate problems we had no means of evaluating before by changingto a radcal.

Example 36.

27− 4

3 Change to radical, denominator is index,negativemeans reciprocal

1

( 273√

)4Evaluate radical

1

(3)4Evaluate exponent

25

1

81Our solution

The largest advantage of being able to change a radical expression into an expo-nential expression is we are now allowed to use all our exponent properties to sim-plify. The following table reviews all of our exponent properties.

Properties of Exponents

aman = am+n (ab)m = ambm a−m =1

am

am

an= am−n

(

a

b

)m

=am

bm

1

a−m= am

(am)n = amn a0 =1(

a

b

)−m

=bm

am

When adding and subtracting with fractions we need to be sure to have acommon denominator. When multiplying we only need to multiply the numera-tors together and denominators together. The following examples show severaldifferent problems, using different properties to simplify the rational exponents.

Example 37.

a2

3 b1

2 a1

6 b1

5 Need common denominator on a ′s(6) and b′s(10)

a4

6 b5

10 a1

6 b2

10 Add exponents on a′s and b′s

a5

6 b7

10 Our Solution

Example 38.

(

x1

3 y2

5

)3

4

Multiply34by each exponent

x1

4y3

10 Our Solution

Example 39.

x2y2

3 · 2x1

2 y5

6

x7

2y0In numerator, need common denominator to add exponents

26

x4

2 y4

6 · 2x1

2 y5

6

x7

2y0Add exponents in numerator, in denominator, y0 =1

2x5

2 y9

6

x7

2

Subtract exponents onx, reduce exponent on y

2x−1y3

2 Negative exponentmoves down to denominator

2y3

2

xOur Solution

Example 40.

25x1

3 y2

5

9x4

5 y− 3

2

− 1

2

Using order of operations, simplify inside parenthesis firstNeed common denominators beforewe can subtract exponents

25x5

15 y4

10

9x12

15 y− 15

10

− 1

2 Subtract exponents,be careful of the negative:

4

10−

(

−15

10

)

=4

10+

15

10=

19

10

(

25x− 7

15 y19

10

9

)− 1

2

Thenegative exponentwill flip the fraction

(

9

25x− 7

15 y19

10

)1

2

The exponent1

2goes on each factor

91

2

251

2 x− 7

30 y19

20

Evaluate 91

2 and 251

2 andmove negative exponent

3x7

30

5y19

20

Our Solution

It is important to remember that as we simplify with rational exponents we areusing the exact same properties we used when simplifying integer exponents. Theonly difference is we need to follow our rules for fractions as well. It may be worthreviewing your notes on exponent properties to be sure your comfortable withusing the properties.

27

Practice - Rational Exponents

Write each expression in radical form.

1) m3

5

3) (7x)3

2

2) (10r)− 3

4

4) (6b)− 4

3

Write each expression in exponential form.

5)1

( 6x√

)3

7)1

( n4√

)7

6) v√

8) 5a√

Simplify. Your answer should contain only positive exponents.

13) yx1

3 ·xy3

2

15) (a1

2b1

2)−1

17)a2b0

3a4

19) uv · u · (v3

2)3

21) (x0y1

3)3

2x0

23)a

3

4b−1 · b7

4

3b−1

25)3y

−

5

4

y−1 · 2y−

1

3

27) (m

3

2n−2

(mn

4

3)−1

)7

4

29)(m2n

1

2)0

n

3

4

31)(x

−

4

3y−

1

3 · y)−1

x

1

3y−2

33)(uv2)

1

2

v−

1

4v2

14) 4v2

3 · v−1

16) (x5

3y−2)0

18)2x

1

2y

1

3

2x

4

3y−

7

4

20) (x ·xy2)0

22) u− 5

4v2 · (u3

2)− 3

2

24)2x−2y

5

3

x−

5

4y−

5

3 · xy

1

2

26)ab

1

3 · 2b−

5

4

4a−

1

2b−

2

3

28)(y

−

1

2)3

2

x

3

2y

1

2

30)y0

(x3

4y−1)1

3

32)(x

1

2y0)−

4

3

y4 · x−2y−

2

3

34) (y

1

3y−2

(x5

3y3)−

3

2

)3

2

28

8.7

Radicals - Mixed Index

Knowing that a radical has the same properties as exponents (written as a ratio)allows us to manipulate radicals in new ways. One thing we are allowed to do isreduce, not just the radicand, but the index as well. This is shown in the fol-lowing example.

Example 41.

x6y28√

Rewrite as raitonal exponent

(x6y2)1

5 Multiply exponents

x6

8 y2

8 Reduce each fraction

x3

4y1

4 All exponents have denominator of 4, this is our new index

x3y4√

Our Solution

What we have done is reduced our index by dividing the index and all the expo-nents by the same number (2 in the previous example). If we notice a commonfactor in the index and all the exponnets on every factor we can reduce bydividing by that common factor. This is shown in the next example

Example 42.

a6b9c1524√

Index and all exponents are divisible by 3

a2b3c58√

Our Solution

We can use the same process when there are coefficients in the problem. We willfirst write the coefficient as an exponential expression so we can divide theexponet by the common factor as well.

Example 43.

8m6n39√

Write 8 as 23

23m6n39√

Index and all exponents are divisible by 3

2m2n3√

Our Solution

29

We can use a very similar idea to also multiply radicals where the index does notmatch. First we will consider an example using rational exponents, then identifythe pattern we can use.

Example 44.

ab23√

a2b4√

Rewrite as rational exponents

(ab2)1

3(a2b)1

4 Multiply exponents

a1

3 b2

3 a2

4 b1

4 Tohave one radical need a commondenominator, 12

a4

12 b8

12 a6

12 b3

12 Write under a single radical with common index, 12

a4b8a6b312√

Add exponents

a10b1112√

Our Solution

To combine the radicals we need a common index (just like the common denomi-nator). We will get a common index by multiplying each index and exponent byan integer that will allow us to build up to that desired index. This process isshown in the next example.

Example 45.

a2b34√

a2b6√

Common index is 12.

Multiply first index and exponents by 3, second by 2

a6b9a4b212√

Add exponents

a10b1112√

Our Solution

Often after combining radicals of mixed index we will need to simplify theresulting radical.

Example 46.

x3y45√

x2y3√

Common index: 15.

Multiply first index and exponents by 3, second by 5

x9y12x10y515√

Add exponents

x19y1715√

Simplify by dividing exponents by index, remainder is left inside

xy x4y215√

Our Solution

Just as with reducing the index, we will rewrite coefficients as exponential expres-sions. This will also allow us to use exponent properties to simplify.

30

Example 47.

4x2y3√

8xy34√

Rewrite 4 as 22 and 8 as 23

22x2y3√

23xy34√

Common index: 12.

Multiply first index and exponents by 4, secondby 3

24x8y429x3y912√

Add exponents (even on the 2)

213x11y1312√

Simplify by dividing exponents by index, remainder is left inside

2y 2x11y12√

Our Solution

If there is a binomial in the radical then we need to keep that binomial togetherthrough the entire problem.

Example 48.

3x(y + z)√

9x(y + z)23√

Rewrite 9 as 32

3x(y + z)√

32x(y + z)23√

Common index: 6.Multiply first groupby 3, second by 2

33x3(y + z)334x2(y + z)46√

Add exponents, keep (y + z) as binomial

37x5(y + z)76√

Simplify,dividing exponent by index, remainder inside

3(y + z) 3x5(y + z)6√

Our Solution

The same process is used for dividing mixed index as with multilpying mixedindex. The only difference is our final answer can’t have a radical over the denom-inator.

Example 49.

x4y3z26√

x7y2z8√ Common index is 24.Multiply first group by 4, secondby 3

x16y12z8

x21y6z324

√

Subtract exponents

x−5y6z524√

Negative exponentmoves to denominator

y6z5

x5

12

√

Can ′t have denominator in radical,need 12x′s, or 7more

y6z5

x5

12

√(

x7

x7

12

√)

Multiply numerator anddenominator by x712√

x7y6z512√

xOur Solution

31

Practice - Radicals of Mixed Index

Reduce the following radicals.

1) 16x4y68√

3) 64x4y6z812√

5)16x2

9y4

6√

7) x6y912√

9) 8x3y68√

11) 8x3y69√

2) 9x2y64√

4)25x3

16x5

4√

6) x9y12z615√

8) 64x8y410√

10) 25y24√

12) 81x8y1216√

Combine the following radicals.

13) 53√

6√

15) x√

7y3√

17) x√

x− 23√

19) x2y5√

xy√

21) xy24√

x2y3√

23) a2bc24√

a2b3c5√

25) a√

a34√

27) b25√

b3√

29) xy3√

x2y3√

31) 9ab34√

9x3yz24√

33) 3xy2z3√

9x3yz24√

35) 27a5(b+ 1)√

81a(b +1)43√

37)a23

√

a4√

39)x2y34

√

xy3√

41)ab3c

√

a2b3c−15√

43)(3x− 1)34

√

(3x− 1)35√

45)(2x+ 1)23

√

(2x+ 1)25√

14) 73√

54√

16) y3√

3z5√

18) 3x4√

y + 4√

20) ab√

2a2b25√

22) a2b35√

a2b4√

24) x2yz36√

x2yz25√

26) x23√

x56√

28) a34√

a23√

30) a3b5√

ab√

32) 2x3y3√

4xy23√

34) a4b3c4√

ab2c3√

36) 8x (y + z)5√

4x2(y + z)23√

38)x23

√

x5√

40)a4b2

5√

ab23√

42)x3y4z95

√

xy−2z√

44)(2+ 5x)23

√

(2 +5x)4√

46)(5− 3x)34

√

(5− 3x)23√

32

8.8

Radicals - Complex Numbers

World View Note: When mathematics was first used the primary purpose wasfor counting. Thus they did not origionally use negatives, zero, fractions or irra-tional numbers. However, the ancient Egyptians quickly developed the need for “apart” and so they made up a new type of number, the ratio or fraction. TheAncient Greeks did not believe in irrational numbers (people were killed forbelieving otherwise). The Mayans of Central America later made up the numberzero when they found use for it as a placeholder. Ancient Chinese Mathematiciansmade up negative numbers when they found use for them.

In mathematics, when the current number system does not provide the tools tosolve the problems the culture is working with, we tend to make up new ways fordealing with the problem that can solve the problem. Throughout history this hasbeen the case with the need for a number that is nothing (0), smaller than zero(negatives), between integers (fractions), and between fractions (irrational num-bers). This is also the case for the square roots of negative numbers. To workwith the square root of negative numbers mathematicians have defined what arecalled imaginary and complex numbers.

Definition of ImaginaryNumbers: i2 = − 1 (thus i = − 1√

)

Examples of imaginary numbers include 3i, − 6i,3

5i and 3i 5

√. A complex

number is one that contains both a real and imaginary part, such as 2+ 5i.

With this definition, the square root of a negative number is no longer undefined.We now are allowed to do basic operations with the square root of negatives.First we will consider exponents on imaginary numbers. We will do this bymanipulating our definition of i2 = − 1. If we multiply both sides of the definitionby i, the equation becomes i3 = − i. Then if we multiply both sides of the equa-tion again by i, the equation becomes i4 = − i2 = − ( − 1) = 1, or simply i4 = 1.Multiplying again by i gives i5 = i. One more time gives i6 = i2 = − 1. And if thispattern continues we see a cycle forming, the exponents on i change we cyclethrough simplified answers of i, − 1, − i, 1. As there are 4 different possibleanswers in this cycle, if we divide the exponent by 4 and consider the remainder,we can simplify any exponent on i by learning just the following four values:

Cyclic Property of Powers of i

i0 =1i = i

i2 =− 1i3 =− i

33

Example 50.

i35 Divide exponent by 4

8R3 Use remainder as exponent on i

i3 Simplify

− i Our Solution

Example 51.

i124 Divide exponent by 4

31R0 Use remainder as exponent on i

i0 Simplify

1 Our Solution

When performing operations (add, subtract, multilpy, divide) we can handel i justlike we handel any other variable. This means when adding and subtracting com-plex numbers we simply add or combine like terms.

Example 52.

(2 +5i) + (4− 7i) Combine like terms 2+ 4 and 5i− 7i

6− 2i Our Solution

It is important to notice what operation we are doing. Students often see theparenthesis and think that means FOIL. We only use FOIL to multiply. Thisproblem is an addition problem so we simply add the terms, or combine liketerms.

For subtraction problems the idea is the same, we need to remember to first dis-tribute the negative onto all the terms in the parentheses.

Example 53.

(4− 8i)− (3− 5i) Distribute the negative

4− 8i− 3+5i Combine like terms 4− 2 and− 8i+ 5i


Addition and subtraction can be combined into one problem.

Example 54.

(5i)− (3+ 8i)+ (− 4 +7i) Distribute the negative

5i− 3− 8i− 4 +7i Combine like terms 5i− 8i + 7i and− 3− 4

− 7 +4i Our Solution

Multiplying with complex numbers is the same as multiplying with variables withone exception, we will want to simplify our final answer so there are no exponentson i.

34

Example 55.

(3i)(7i) Multilpy coefficients and i′s

21i2 Simplify i2 =− 1

21(− 1) Multiply

− 21 Our Solution

Example 56.

5i(3i− 7) Distribute

15i2− 35i Simplify i2 =− 1

15(− 1)− 35i Multiply

− 15− 35i Our Solution

Example 57.

(2− 4i)(3 +5i) FOIL

6+ 10i− 12i− 20i2 Simplify i2 =− 1

6+ 10i− 12i− 20(− 1) Multiply

6+ 10i− 12i+ 20 Combine like trems 6+ 20 and 10i− 12i


Example 58.

(3i)(6i)(2− 3i) Multiply first twomonomials

18i2(2− 3i) Distribute

36i2− 48i3 Simplify i2 =− 1 and i3 =− i

36(− 1)− 48(− i) Multiply

− 36+ 48i Our Solution

Remember when squaring a binomial we either have to FOIL or use our shortcutto square the first, twice the product and square the last. The next example usesthe shortcut

Example 59.

(4− 5i)2 Use perfect square shortcut

42 = 16 Square the first

2(4)(− 5i)=− 40i Twice the product

(5i)2 = 25i2 = 25(− 1)=− 25 Square the last, simplify i2 =− 1

16− 40i− 25 Combine like terms

− 9− 40i Our Solution

Dividing with complex numbers also has one thing we need to be careful of. If i is

− 1√

, and it is in the denominator of a fraction, then we have a radical in the

35

denominator! This means we will want to rationalize our denominator so thereare no i’s. This is done the same way we rationalized denoinators with squareroots.

Example 60.

7 +3i

− 5iJust amonomial in denominator,multiply by i

7+ 3i− 5i

(

i

i

)

Distribute i in numerator

7i + 3i2

− 5i2Simplify i2 =− 1

7i+ 3(− 1)

− 5(− 1)Multiply

7i− 3

5Our Solution

The solution for these problems can be written several different ways, for example− 3+ 7i

5or

− 3

5+

7

5i, The author has elected to use the solution as written, but it is

important to express your answer in the form your instructor prefers.

Example 61.

2− 6i

4+ 8iBinomial in denominator,multiply by conjugate, 4− 8i

2− 6i

4 +8i

(

4− 8i

4− 8i

)

FOIL in numerator,denominator is a difference of squares

8− 16i− 24i + 48i2

16− 64i2Simplify i2 =− 1

8− 16i− 24i + 48(− 1)

16− 64(− 1)Multiply

8− 16i− 24i− 48

16+ 64Combine like terms 8− 48 and− 16i− 24i and 16+ 64

− 40− 40i

80Reduce,divide each termby 40

− 1− i

2Our Solution

36

Practice - Complex Numbers

Simplify.

1) 3− (− 8+4i)

3) (7i)− (3− 2i)

5) (− 6i)− (3+ 7i)

7) (3− 3i)+ (− 7− 8i)

9) (i)− (2+ 3i)− 6

11) (6i)(− 8i)

13) (− 5i)(8i)

15) (− 7i)2

17) (6+ 5i)2

19) (− 7− 4i)(− 8+6i)

21) (− 4+ 5i)(2− 7i)

23) (− 8− 6i)(− 4+2i)

25) (1+ 5i)(2+ i)

27)− 9+ 5i

i

29)− 10− 9i

6i

31)− 3− 6i

4i

33)10− i

− i

35)4i

− 10+ i

37)8

− 10+ i

39)7

10− 7i

41)5i

− 6− i

2) (3i)− (7i)

4) 5+ (− 6− 6i)

6) (− 8i)− (7i)− (5− 3i)

8) (− 4− i)+ (1− 5i)

10) (5− 4i)+ (8− 4i)

12) (3i)(− 8i)

14) (8i)(− 4i)

16) (− i)(7i)(4− 3i)

18) (8i)(− 2i)(− 2− 8i)

20) (3i)(− 3i)(4− 4i)

22) − 8(4− 8i)− 2(− 2− 6i)

24) (− 6i)(3− 2i)− (7i)(4i)

26) (− 2+ i)(3− 5i)

28)− 3+ 2i

− 3i

30)− 4+ 2i

3i

32)− 5+ 9i

9i

34)10

5i

36)9i

1− 5i

38)4

4+ 6i

40)9

− 8− 6i

42)8i

6− 7i

37

Answers - Square Roots

1) 7 5√

2) 5 5√

3) 6

4) 14

5) 2 3√

6) 6 2√

7) 6 3√

8) 20 2√

9) 48 2√

10) 56 2√

11) − 112 2√

12) − 21 7√

13) 8 3n√

14) 7 7b√

15) 14v

16) 10n n√

17) 6x 7√

18) 10a 2a√

19) − 10k2

20) − 20p2 7√

21) − 56x2

22) − 16 2n√

23) − 30 m√

24) 32p 7√

25) 3xy 5√

26) 6b2a 2a√

27) 4xy xy√

28) 16a2b 2√

29) 8x2y2 5√

30) 16m2n 2n√

31) 24y 5x√

32) 56 2mn√

33) 35xy 5y√

34) 12xy 2√

35) − 12u 5uv√

36) − 30y2x 2x√

37) − 48x2z2y 5√

38) 30a2c 2b√

39) 8j2 5hk√

40) − 4yz 2xz√

41) − 12p 6mn√

42) − 32p2m 2q√

Answers - Higher Roots

1) 5 53√

2) 5 33√

3) 5 63√

4) 5 23√

5) 5 73√

6) 2 33√

7) − 8 64√

8) − 16 34√

9) 12 74√

10) 6 34√

11) − 2 74√

12) 15 34√

13) 3 8a24√

14) 2 4n34√

15) 2 7n35√

16) − 2 3x45√

17) 2p 75√

18) 2x 46√

19) − 6 7r7√

20) − 16b 3b7√

21) 4v2 6v3√

22) 20a2 23√

23) − 28n2 53√

24) − 8n2

25) − 3xy 5x23√

26) 4uv u23√

27) − 2xy 4xy3√

28) 10ab ab23√

29) 4xy2 4x3√

30) 3xy2 73√

31) − 21xy2 3y3√

32) − 8y2 7x2y23√

33) 10v2 3u2v23√

34) − 40 6xy3√

35) − 12 3ab23√

36) 9y 5x3√

37) − 18m2np2 2m2p3√

38) − 12mpq 5p34√

39) 18xy 8xy3z24√

40) − 18b2a 5ac4√

41) 14j2k2h 8h24√

42) − 18xz 4x3yz34√

38

Answers - Add and Subtract Radicals

1) 6 5√

2) − 3 6√

− 5 3√

3) − 3 2√

+6 5√

4) − 5 6√

− 3√

5) − 5 6√

6) − 3 3√

7) 3 6√

+ 5 5√

8) − 5√

+ 3√

9) − 8 2√

10) − 6 6√

+9√

3

11) − 3 6√

+ 3√

12) − 2 5√

− 6 6√

13) − 2 2√

14) 8 5√

− 3√

15) 5 2√

16) − 9 3√

17) − 3 6√

− 3√

18) 3 2√

+ 3 6√

19) − 12 2√

+2 5√

20) − 3 2√

21) − 4 6√

+4 5√

22) − 5√

− 3 6√

23) 8 6√

− 9 3√

+4 2√

24) − 6√

− 10 3√

25) 2 23√

26) 6 53√

− 3 33√

27) − 34√

28) 10 44√

29) 24√

− 3 34√

30) 5 64√

+ 2 44√

31) 3 44√

32) − 6 34√

+2 64√

33) 2 24√

+ 34√

+ 6 44√

34) − 2 34√

− 9 54√

− 3 24√

35) 65√

− 6 25√

36) 10 65√

− 9 55√

37) 4 55√

− 4 65√

38) − 11 27√

− 2 57√

39) − 4 46√

− 6 56√

− 4 26√

40) 37√

− 6 67√

+3 57√

Answers - Multiply and Divide Radicals

1) − 48 5√

2) − 25 6√

3) 6m 5√

4) − 25r2 2r√

5) 2x2 x3√

6) 6a2 5a3√

7) 2 3√

+ 2 6√

8) 5 2√

+ 2 5√

9) − 45 5√

− 10 15√

10) 25 6r√

+ 20r2 15√

11) 25n 10√

+ 10 5√

12) 5 3√

− 9 5√

13) − 2− 4 2√

14) 16− 9 3√

15) 15− 11 5√

16) 30+8 3√

+5 15√

+4 5√

39

17) 6a+ a 10√

+6a 6√

+ 2a 15√

18) − 4p 10√

+ 50 p√

19) 63+ 32 3√

20) − 10 m√

+ 25 2√

+ 2m√

− 5

21)3

√

25

22)15

√

4

23)1

20

24) 2

25)15

√

3

26)10

√

15

27)4 3√

9

28)4 5√

5

29)5 3xy√

12y2

30)4 3x√

15y2x

31)6p

√

3

32)2 5n√

5

33)10

3√

5

34)15

3√

4

35)10

3√

8

36)84

√

8

37)5 10r24√

2

38)4n24

√

mn

Answers - Rationalize Denominators

1)4+ 2 3

√

3

2)− 4+ 3

√

12

3)2+ 3

√

5

4)3

√− 1

4

5)2 13√

− 5 65√

52

6)85

√+ 4 17

√

68

7)6

√− 9

3

8)30

√− 2 3

√

18

9)2 5√

+ 15p

50

10)5 19√

− x 57√

38

11)3− 4m 6

√

30m

12)6 7r√

− 42r√

14

13)15 5

√− 5 2

√

43

14)− 5 3

√+20 5

√

77

15)10− 2 2

√

23

16)2 3√

+ 2√

2

17)− 12− 9 3

√

11

18) − 2 2√

− 4

19) 3− 5√

20)5

√− 3

√

2

21) 1− 2√

22)16 3

√+ 4 5

√

43

23)5

n2− 5

24)− 5n 3n

√+n3 6n

√

25− 2n4

25)4p

3− 5p2

26)5x2 + 3x2 5x

√

5− 9x

27)20− 4x 5

√

25− 5x2

40

28)2b

− 2b2− 1

29)10− 5r 5r

√

4− 5r3

30)5a 3

√− 20

3a3− 16a

31)− 25v + 15 v

√

25v2− 9v

32)8 2√

− 4 n√

8n −n2

33)24− 4 6

√+ 9 2

√− 3 3

√

15

34)5+ 5 3

√+ 4 5

√+ 4 15

√

10

35)− 1+ 5

√

4

36)2 5√

− 5 2√

−10+ 5 10√

30

37)− 5 2

√+ 10− 3

√+ 6√

5

38)− 4 10

√− 16 2

√− 2 15

√− 8 3

√

11

39)8+ 3 6

√

10

40)4 3√

+ 6√

− 4 5√

− 10√

14

41)4 3√

− 15√

− 4 2√

+ 10√

11

42)− 2 10

√+ 6 6

√+ 15

√− 9

22

43)20− 8x 5x

√+ 10x 2

√− 4x2 10x

√

25−20x3

44)12p2 + 3p4 2

√+ 4p 5p

√+ p3 10p

√

16− 2p4

45)10m 3

√+ 6m2− 5m2 2

√−m3 6

√

25− 3m2

46)− 16v + 20v v

√+ 8v 2

√− 10v 2v

√

16− 25v

47)− 2b− 2b3 2

√+5 2b

√+ 10b2 b

√

1− 2b4

48)− 4n −n 3

√

− 3−n2

49)8+ 10 3x

√− 4 2x

√− 5x 6

√

16x − 75x2

50)9a +15a2 3

√+ 6a 2

√+10a2 6

√

9− 75a2

51)3p − 4p p

√+ p√

p− p2

52)− 8+ 4 3

√+ 2 5x

√− 15x

√

x

Answers - Rational Exponents

1) ( m5√

)3

2)1

( 10r4√

)3

3) ( 7x√

)3

4)1

( 6v3√

)4

5) (6x)− 3

2

6) v1

2

7) n− 7

4

8) (5a)1

2

9) 4

10) 2

11) 8

12)1

1000

13) x4

3y5

2

14)4

v

1

3

15)1

a

1

2b

1

2

16) 1

17)1

3a2

18)y

25

12

x

5

6

19) u2v11

2

20) 1

21) y1

2

22)v2

u

7

2

23)b

7

4a

3

4

3

24)2y

17

6

x

7

4

25)3y

1

12

2

26)a

3

2

2b

1

4

27)m

35

8

n

7

6

28)1

y

5

4x

3

2

29)1

n

3

4

30)y

1

3

x

1

4

41

31) xy4

3

32)x

4

3

y

10

3

33)u

1

2

v

3

4

34) x15

4 y17

4

Answers - Mixed Index

1) 4x2y34√

2) 3xy3√

3) 8x2y3z46√

4)5

4x

√

5)36xy3

√

3y

6) x3y4z25√

7) x2y34√

8) 8x4y25√

9) x3y2z4√

10) 5y√

11) 2xy23√

12) 3x2y34√

13) 54006√

14) 30012512√

15) 49x3y26√

16) 27y5z515√

17) x3(x− 2)26√

18) 3x(y +4)24√

19) x9y710√

20) 4a9b910√

21) x11y1012√

22) a18b1720√

23) a18b17c1420√

24) x22y11z2730√

25) a a4√

26) x x√

27) b b910√

28) a a512√

29) xy xy56√

30) a ab710√

31) 3a2b ab4√

32) 2xy2 2x5y6√

33) 3xy xy4√

34) a2b2c2 a2b c26√

35) 9a2(b +1) 243a5(b +1)56√

36) 4x(y + z)3 2x(y + z)6√

37) a512√

38) x715√

39) x2y512√

40)a7b11

15√

b

41) ab9c710√

42) yz xy8z310√

43) (3x− 1)320√

44) (2+ 5x)512√

42

45) (2x +1)415√

Answers - Complex Numbers

1) 11− 4i

2) − 4i

3) − 3+ 9i

4) − 1− 6i

5) − 3− 13i

6) 5− 12i

7) − 4− 11i

8) − 3− 6i

9) − 8− 2i

10) 13− 8i

11) 48

12) 24

13) 40

14) 32

15) − 49

16) 28− 21i

17) 11+ 60i

18) − 32− 128i

19) 80− 10i

20) 36− 36i

21) 27+ 38i

22) − 28+ 76i

23) 44+8i

24) 16− 18i

25) − 3+ 11i

26) − 1+ 13i

27) 9i + 5

28)− 3i− 2

3

29)10i − 9

6

30)4i + 2

3

31)3i− 6

4

32)5i + 9

9

33) 10i +1

34) − 2i

35)− 40i + 4

101

36)9i − 45

26

37)56+ 48i

85

38)4− 6i

13

39)70+ 49i

149

40)− 36+27i

50

41)− 30i− 5

37

42)48i −56

85

43

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